Server: Netscape-Enterprise/2.0a
Date: Wed, 13 Nov 1996 23:15:08 GMT
Content-type: text/html

<HEAD>
<TITLE>Homework #1 Solutions</TITLE>
</HEAD>
<BODY>

<h2>Solutions to Homework #1</H2>
<p>
<p>

 <b> Question 2.1</b> Cost effectiveness is performance per dollar of cost,
and the performance will depend on the workload choice.
<P>
Cost effectiveness can be measured in the numbers of executions per second
per dollar of machine cost:
<P>
executions per second per dollar = <IMG  ALIGN=MIDDLE ALT="" SRC="img1.gif">  <BR>
<P>
executions per second per dollar for S1  = <IMG  ALIGN=MIDDLE ALT="" SRC="img2.gif"> <BR>
<P>
executions per second per dollar for S2 dollars = <IMG  ALIGN=MIDDLE ALT="" SRC="img3.gif"> <BR>
<P>
So System 2 is more cost effective.  By a ratio of  <IMG  ALIGN=MIDDLE ALT="" SRC="img4.gif">
<P>
<hr>

<b> Question 2.2</b> We can use the same method as the previous part, but
the execution time is the sum of the two execution times.
<P>
executions per second per dollar for S1 = <IMG  ALIGN=MIDDLE ALT="" SRC="img5.gif"> <BR>
<P>
executions per second per dollar for S2 = <IMG  ALIGN=MIDDLE ALT="" SRC="img6.gif"> <BR>
<P>
So System 1 is slightly more cost effective, by a factor of
<IMG  ALIGN=MIDDLE ALT="" SRC="img7.gif"> <BR>
<P>
<hr>

<b> Question 2.3</b> We do this problem by finding the amount of time that
program 2 can be run in an hour and using that for executions per second,
the throughput measure.
<P>
executions of P2 per hour = <IMG  ALIGN=MIDDLE ALT="" SRC="zz1.gif">
<BR>
<P>
executions of P2 per hour on S1 = <IMG  ALIGN=MIDDLE ALT="" SRC="zz2.gif">
<BR>
<P>
executions of P2 per hour on S2 = <IMG  ALIGN=MIDDLE ALT="" SRC="zz3.gif">
<BR>
<P>
Hence we can conclude that S2 has higher throughput. <BR>
<P>
Cost per run per hour for P2 in dollars: <BR>
<P>
For S1 =  <IMG  ALIGN=MIDDLE ALT="" SRC="zz4.gif"> = 18.76 <BR>
<P>
For S2 =  <IMG  ALIGN=MIDDLE ALT="" SRC="zz5.gif"> = 23.07 <BR>
<P>
Hence S1 is more cost effective for this workload.  <BR>
<P> 

<hr>
<b> Question 2.27</b>
<P>
<IMG  ALIGN=MIDDLE ALT="" SRC="img11.gif"> <BR>
<P>
Harmonic Mean (HM) = <IMG  ALIGN=MIDDLE ALT="" SRC="img12.gif"> <BR>
<P>
mean of times = <IMG  ALIGN=MIDDLE ALT="" SRC="img13.gif"><BR>
<P>
mean of times = <IMG  ALIGN=MIDDLE ALT="" SRC="img14.gif"><BR>
<P>
Therefore, HM = <IMG  ALIGN=MIDDLE ALT="" SRC="img15.gif"><BR>
<P>

<hr>
<b> State machine</b>
<P>
A <a href="match.ps">schematic</a> of a possible solution to the state
machine problem.

The <a href="sim.ps">waveform</a> of the simulation of this schematic.

<p>
<BR> <HR>
<P><ADDRESS>
<I>Course Account ece552 <BR>
Thu Sep 12 15:55:17 CDT 1996</I>
</ADDRESS>
</BODY>
